Good day to everyone!
I just have a quick question regarding the greatest common divisor function.
Say I have $\gcd(m,n^2)=1$. Does it follow that $\gcd(m,n)=1$?
Here is my attempt at a proof:
Since $\gcd(m,n^2)=1$, there exist integers $r$ and $s$ such that
$$rm + sn^2 = 1.$$
It follows that there exist integers $r$ and $t=sn$ such that
$$rm + tn = 1.$$
Therefore, $\gcd(m,n)=1$.
But it can't be that easy! Is my proof correct?
Thanks!
Yes, the proof is that easy if you are allowed to use the Bezout identity. Else, more generally
${\bf Lemma}\quad \begin{align}a\mid A\\ b\mid B\end{align}\,\Rightarrow\,(a,b)\mid (A,B)\ \ $
${\bf Proof}\ \ \ \begin{align}(a,b)\mid a\mid A\\ (a,b)\mid b\mid B\end{align}\,\Rightarrow\, (a,b)\mid A,B\,\Rightarrow\,(a,b)\mid (A,B)$
Corollary $\ $ If $\,A,B$ are coprime so too are any respective factors $\,a,b\,$ (i.e. $\,a\mid A,\,b\mid B)$
Remark $\ $ Unlike the proof using Bezout, the above proof works in any domain where gcds exist, e.g. non-PID UFDs like $\,\Bbb Z[x]\,$ and $\,F[x,y]\,$ where there is no Bezout identity for the gcd.